ABC is a triangle inscribed in a circle, AC being the diameter of the circle. The length of AC is as much more than the length of BC as the length of BC is more than the length of AB. Find AC:AB.

							


Dear Vipul,
Solution:- Given, AC = a BC and BC = a AB. We want to find AC: AB  = a2
Since, AC is the diameter of the circle, it subtends 900 angle at B. Let, AC = 'd' and angle C be 'x'. Then, 
AB = d sinx , and BC = d cosx
So, AC = a BC will give 1 = a cosx     or    cosx = 1/a

And, BC = a AB will give cos x = a sinx 
Since, cos x = 1/a, sin x = √(a2 -1) / a
So, 1 = a √(a2 -1)          or         a4 - a2 -1 = 0
On solving quad. eq., we'll get - a2 = (√5 +1) / 2  [ANS]

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ΔABC is an right angle triangle as Ac is the diameter of the circle. 
Now it is given that AC-BC=BC-AB
AC+AB=2BC
AC=2BC-AB............(1)
AB²+BC²=AC². .........(Pythagoras theorem) 
AB²+BC²=(2BC-AB)²............(using eqn.  1)
AB²+BC²=4BC²+AB²-4*BC*AB
BC²-4BC²=-4*BC*AB
3BC²=4*BC*AB
3BC²/4BC=AB
3/4BC=AB........(2)
AB²+BC²=AC²
(3/4BC)²+BC²=AC².......(Using eqn. 2)
9/16BC²+BC²=AC²
9/16BC²+16/16BC²=AC²
25/16BC²=AC²
5/4BC=AC........(3)
AC:AB=AC/AB
(5/4BC)/(3/4BC)..........(Using eqn.  2 &3)
5/3=AC/AB=AC:AB